Integrating Products and Powers of sinx and cosx

Learning Outcomes

Solve integration problems involving products and powers of $\sin{x}$ and $\cos{x}$

A key idea behind the strategy used to integrate combinations of products and powers of $\sin{x}$ and $\cos{x}$ involves rewriting these expressions as sums and differences of integrals of the form $\displaystyle\int\sin^{j}x\cos{x}dx$ or ${\displaystyle\int}{\cos}^{j}x\sin{x}dx$ . After rewriting these integrals, we evaluate them using u-substitution.

Before describing the general process in detail, let’s take a look at the following examples.

Example: Integrating ${\displaystyle\int}{\cos}^{j}x\sin{x}dx$

Evaluate ${\displaystyle\int}{\cos}^{3}x\sin{x}dx$ .

Show Solution

Use $u$ -substitution and let $u=\cos{x}$ . In this case, $du=\text{-}\sin{x}dx$ . Thus,

\begin{array}{cc} \int \cos^{3} x \sin x d x & = - \int u^{3} d u \\ = - \frac{1}{4} u^{4} + C \\ = - \frac{1}{4} \cos^{4} x + C . \end{array}

$\begin{array}{cc}{\displaystyle\int}{\cos}^{3}x\sin{x}dx\hfill & =\text{-}{\displaystyle\int}{u}^{3}du\hfill \\ \hfill & =-\frac{1}{4}{u}^{4}+C\hfill \\ \hfill & =-\frac{1}{4}{\cos}^{4}x+C.\hfill \end{array}$

try it

Evaluate ${\displaystyle\int}{\sin}^{4}x\cos{x}dx$ .

Hint

Let $u=\sin{x}$ .

Show Solution

$\frac{1}{5}{\sin}^{5}x+C$

Watch the following video to see the worked solution to the above Try It

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You can view the transcript for this segmented clip of “3.2 Trigonometric Integrals” here (opens in new window).

In addition to the technique of

u -

$u-$ substitution, the problems in this section and the next make frequent use of the Pythagorean Identity and its implications for how to rewrite trigonometric functions in terms of other trigonometric functions. We briefly review the relationships between these functions below.

Recall: The Pythagorean Identity

For any angle $x$ :

$\sin^2 x + \cos^2 x = 1$

Subtracting by $\sin^2 x$ allows a square power of cosine in terms of sine:
$\cos^2 x = 1-\sin^2 x$

Subtracting instead by $\cos^2 x$ allows a square power of sine to be written in terms of cosine:

$\sin^2 x = 1 -\cos^2 x$

Example: Integrating ${\displaystyle\int}{\cos}^{j}x{\sin}^{k}xdx$ Where k is Odd

Evaluate ${\displaystyle\int}{\cos}^{2}x{\sin}^{3}xdx$ .

Show Solution

To convert this integral to integrals of the form ${\displaystyle\int}{\cos}^{j}x\sin{x}dx$ , rewrite ${\sin}^{3}x={\sin}^{2}x\sin{x}$ and make the substitution ${\sin}^{2}x=1-{\cos}^{2}x$ . Thus,

\begin{array}{ccc} \int \cos^{2} x \sin^{3} x d x & = \int \cos^{2} x (1 - \cos^{2} x) \sin x d x & Let u = \cos x; then d u = - \sin x d x . \\ = - \int u^{2} (1 - u^{2}) d u \\ = \int (u^{4} - u^{2}) d u \\ = \frac{1}{5} u^{5} - \frac{1}{3} u^{3} + C \\ = \frac{1}{5} \cos^{5} x - \frac{1}{3} \cos^{3} x + C . \end{array}

$\begin{array}{ccc}{\displaystyle\int}{\cos}^{2}x{\sin}^{3}xdx\hfill & ={\displaystyle\int}{\cos}^{2}x\left(1-{\cos}^{2}x\right)\sin{x}dx\hfill & \text{Let }u=\cos{x};\text{then }du=\text{-}\sin{x}dx.\hfill \\ \hfill & =\text{-}{\displaystyle\int}{u}^{2}\left(1-{u}^{2}\right)du\hfill & \hfill \\ \hfill & ={\displaystyle\int}\left({u}^{4}-{u}^{2}\right)du\hfill & \hfill \\ \hfill & =\frac{1}{5}{u}^{5}-\frac{1}{3}{u}^{3}+C\hfill & \hfill \\ \hfill & =\frac{1}{5}{\cos}^{5}x-\frac{1}{3}{\cos}^{3}x+C.\hfill & \hfill \end{array}$

try it

Evaluate ${\displaystyle\int}{\cos}^{3}x{\sin}^{2}xdx$ .

Hint

Write ${\cos}^{3}x={\cos}^{2}x\cos{x}=\left(1-{\sin}^{2}x\right)\cos{x}$ and let $u=\sin{x}$ .

Show Solution

$\frac{1}{3}{\sin}^{3}x-\frac{1}{5}{\sin}^{5}x+C$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for this segmented clip of “3.2 Trigonometric Integrals” here (opens in new window).

In the next example, we see the strategy that must be applied when there are only even powers of $\sin{x}$ and $\cos{x}$ . For integrals of this type, the identities

\sin^{2} x = \frac{1}{2} - \frac{1}{2} \cos (2 x) = \frac{1 - \cos (2 x)}{2}

${\sin}^{2}x=\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)=\frac{1-\cos\left(2x\right)}{2}$

and

\cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos (2 x) = \frac{1 + \cos (2 x)}{2}

${\cos}^{2}x=\frac{1}{2}+\frac{1}{2}\cos\left(2x\right)=\frac{1+\cos\left(2x\right)}{2}$

are invaluable. These identities are sometimes known as and they may be derived from the double-angle identity $\cos\left(2x\right)={\cos}^{2}x-{\sin}^{2}x$ and the Pythagorean identity ${\cos}^{2}x+{\sin}^{2}x=1$ .

example: Integrating an Even Power of $\sin{x}$

Evaluate ${\displaystyle\int}{\sin}^{2}xdx$ .

Show Solution

To evaluate this integral, let’s use the trigonometric identity ${\sin}^{2}x=\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)$ . Thus,

\begin{array}{cc} \int \sin^{2} x d x & = \int (\frac{1}{2} - \frac{1}{2} \cos (2 x)) d x \\ = \frac{1}{2} x - \frac{1}{4} \sin (2 x) + C . \end{array}

$\begin{array}{cc}{\displaystyle\int}{\sin}^{2}xdx\hfill & ={\displaystyle\int}\left(\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)\right)dx\hfill \\ \hfill & =\frac{1}{2}x-\frac{1}{4}\sin\left(2x\right)+C.\hfill \end{array}$

try it

Evaluate $\displaystyle\int {\cos}^{2}xdx$ .

Hint

\cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos (2 x)

${\cos}^{2}x=\frac{1}{2}+\frac{1}{2}\cos\left(2x\right)$

Show Solution

$\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)+C$

Try It

The general process for integrating products of powers of $\sin{x}$ and $\cos{x}$ is summarized in the following set of guidelines.

Problem-Solving Strategy: Integrating Products and Powers of sin x and cos x

To integrate ${\displaystyle\int}{\cos}^{j}x{\sin}^{k}xdx$ use the following strategies:

If $k$ is odd, rewrite ${\sin}^{k}x={\sin}^{k - 1}x\sin{x}$ and use the identity ${\sin}^{2}x=1-{\cos}^{2}x$ to rewrite ${\sin}^{k - 1}x$ in terms of $\cos{x}$ . Integrate using the substitution $u=\cos{x}$ . This substitution makes $du=\text{-}\sin{x}dx$ .
If $j$ is odd, rewrite ${\cos}^{j}x={\cos}^{j - 1}x\cos{x}$ and use the identity ${\cos}^{2}x=1-{\sin}^{2}x$ to rewrite ${\cos}^{j - 1}x$ in terms of $\sin{x}$ . Integrate using the substitution $u=\sin{x}$ . This substitution makes $du=\cos{x}dx$ . (Note: If both $j$ and $k$ are odd, either strategy 1 or strategy 2 may be used.)
If both $j$ and $k$ are even, use ${\sin}^{2}x=\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)$ and ${\cos}^{2}x=\frac{1}{2}+\frac{1}{2}\cos\left(2x\right)$ . After applying these formulas, simplify and reapply strategies 1 through 3 as appropriate.

Example: Integrating $\displaystyle\int {\cos}^{j}x{\sin}^{k}xdx$ where k is Odd

Evaluate ${\displaystyle\int}{\cos}^{8}x{\sin}^{5}xdx$ .

Show Solution

Since the power on $\sin{x}$ is odd, use strategy 1. Thus,

\begin{array}{cccc} \int \cos^{8} x \sin^{5} x d x & = \int \cos^{8} x \sin^{4} x \sin x d x & Break off \sin x . \\ = \int \cos^{8} x {(\sin^{2} x)}^{2} \sin x d x & Rewrite \sin^{4} x = {(\sin^{2} x)}^{2} . \\ = \int \cos^{8} x {(1 - \cos^{2} x)}^{2} \sin x d x & Substitute \sin^{2} x = 1 - \cos^{2} x . \\ = \int u^{8} {(1 - u^{2})}^{2} (- d u) & Let u = \cos x and d u = - \sin x d x . \\ = \int (- u^{8} + 2 u^{10} - u^{12}) d u & Expand . \\ = - \frac{1}{9} u^{9} + \frac{2}{11} u^{11} - \frac{1}{13} u^{13} + C & Evaluate the integral . \\ = - \frac{1}{9} \cos^{9} x + \frac{2}{11} \cos^{11} x - \frac{1}{13} \cos^{13} x + C . & Substitute u = \cos x . \end{array}

$\begin{array}{cccc}\hfill {\displaystyle\int}{\cos}^{8}x{\sin}^{5}xdx& ={\displaystyle\int}{\cos}^{8}x{\sin}^{4}x\sin{x}dx\hfill & & \text{Break off }\sin{x}.\hfill \\ & ={\displaystyle\int}{\cos}^{8}x{\left({\sin}^{2}x\right)}^{2}\sin{x}dx\hfill & & \text{Rewrite }{\sin}^{4}x={\left({\sin}^{2}x\right)}^{2}.\hfill \\ & ={\displaystyle\int}{\cos}^{8}x{\left(1-{\cos}^{2}x\right)}^{2}\sin{x}dx\hfill & & \text{Substitute }{\sin}^{2}x=1-{\cos}^{2}x.\hfill \\ & ={\displaystyle\int}{u}^{8}{\left(1-{u}^{2}\right)}^{2}\left(\text{-}du\right)\hfill & & \text{Let }u=\cos{x}\text{ and }du=\text{-}\sin{x}dx.\hfill \\ & ={\displaystyle\int}\left(\text{-}{u}^{8}+2{u}^{10}-{u}^{12}\right)du\hfill & & \text{Expand}.\hfill \\ & =-\frac{1}{9}{u}^{9}+\frac{2}{11}{u}^{11}-\frac{1}{13}{u}^{13}+C\hfill & & \text{Evaluate the integral}.\hfill \\ & =-\frac{1}{9}{\cos}^{9}x+\frac{2}{11}{\cos}^{11}x-\frac{1}{13}{\cos}^{13}x+C.\hfill & & \text{Substitute }u=\cos{x}.\hfill \end{array}$

Example: Integrating ${\displaystyle\int}{\cos}^{j}x{\sin}^{k}xdx$ where k and j are Even

Evaluate ${\displaystyle\int}{\sin}^{4}xdx$ .

Show Solution

Since the power on $\sin{x}$ is even $\left(k=4\right)$ and the power on $\cos{x}$ is even $\left(j=0\right)$ , we must use strategy 3. Thus,

\begin{array}{cccc} \int \sin^{4} x d x & = \int {(\sin^{2} x)}^{2} d x & Rewrite \sin^{4} x = {(\sin^{2} x)}^{2} . \\ = \int {(\frac{1}{2} - \frac{1}{2} \cos (2 x))}^{2} d x & Substitute \sin^{2} x = \frac{1}{2} - \frac{1}{2} \cos (2 x) . \\ = \int (\frac{1}{4} - \frac{1}{2} \cos (2 x) + \frac{1}{4} \cos^{2} (2 x)) d x & Expand {(\frac{1}{2} - \frac{1}{2} \cos (2 x))}^{2} . \\ = \int (\frac{1}{4} - \frac{1}{2} \cos (2 x) + \frac{1}{4} (\frac{1}{2} + \frac{1}{2} \cos (4 x))) d x . \end{array}

$\begin{array}{cccc}\hfill {\displaystyle\int}{\sin}^{4}xdx& ={\displaystyle\int}{\left({\sin}^{2}x\right)}^{2}dx\hfill & & \text{Rewrite }{\sin}^{4}x={\left({\sin}^{2}x\right)}^{2}.\hfill \\ & ={\displaystyle\int}{\left(\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)\right)}^{2}dx\hfill & & \text{Substitute }{\sin}^{2}x=\frac{1}{2}-\frac{1}{2}\cos\left(2x\right).\hfill \\ & ={\displaystyle\int}\left(\frac{1}{4}-\frac{1}{2}\cos\left(2x\right)+\frac{1}{4}{\cos}^{2}\left(2x\right)\right)dx\hfill & & \text{Expand }{\left(\frac{1}{2}-\frac{1}{2}\cos\left(2x\right)\right)}^{2}.\hfill \\ & ={\displaystyle\int}\left(\frac{1}{4}-\frac{1}{2}\cos\left(2x\right)+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\cos\left(4x\right)\right)\right)dx.\hfill & & \end{array}$

Since ${\cos}^{2}\left(2x\right)$ has an even power, substitute ${\cos}^{2}\left(2x\right)=\frac{1}{2}+\frac{1}{2}\cos\left(4x\right)\text{:}$

\begin{array}{cc} = \int (\frac{3}{8} - \frac{1}{2} \cos (2 x) + \frac{1}{8} \cos (4 x)) d x & Simplify . \\ = \frac{3}{8} x - \frac{1}{4} \sin (2 x) + \frac{1}{32} \sin (4 x) + C & Evaluate the integral . \end{array}

$\begin{array}{cc}={\displaystyle\int}\left(\frac{3}{8}-\frac{1}{2}\cos\left(2x\right)+\frac{1}{8}\cos\left(4x\right)\right)dx\hfill & \text{Simplify}.\hfill \\ =\frac{3}{8}x-\frac{1}{4}\sin\left(2x\right)+\frac{1}{32}\sin\left(4x\right)+C\hfill & \text{Evaluate the integral}.\hfill \end{array}$

try it

Evaluate ${\displaystyle\int}{\cos}^{3}xdx$ .

Hint

Use strategy 2. Write ${\cos}^{3}x={\cos}^{2}x\cos{x}$ and substitute ${\cos}^{2}x=1-{\sin}^{2}x$ .

Show Solution

$\sin{x}-\frac{1}{3}{\sin}^{3}x+C$

try it

Evaluate ${\displaystyle\int}{\cos}^{2}\left(3x\right)dx$ .

Hint

Use strategy 3. Substitute ${\cos}^{2}\left(3x\right)=\frac{1}{2}+\frac{1}{2}\cos\left(6x\right)$

Show Solution

$\frac{1}{2}x+\frac{1}{12}\sin\left(6x\right)+C$

In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include $\sin\left(ax\right)$ , $\sin\left(bx\right)$ , $\cos\left(ax\right)$ , and $\cos\left(bx\right)$ . These integrals are evaluated by applying trigonometric identities, as outlined in the following rule.

Rule: Integrating Products of Sines and Cosines of Different Angles

To integrate products involving $\sin\left(ax\right)$ , $\sin\left(bx\right)$ , $\cos\left(ax\right)$ , and $\cos\left(bx\right)$ , use the substitutions

\sin (a x) \sin (b x) = \frac{1}{2} \cos ((a - b) x) - \frac{1}{2} \cos ((a + b) x)

$\sin\left(ax\right)\sin\left(bx\right)=\frac{1}{2}\cos\left(\left(a-b\right)x\right)-\frac{1}{2}\cos\left(\left(a+b\right)x\right)$

\sin (a x) \cos (b x) = \frac{1}{2} \sin ((a - b) x) + \frac{1}{2} \sin ((a + b) x)

$\sin\left(ax\right)\cos\left(bx\right)=\frac{1}{2}\sin\left(\left(a-b\right)x\right)+\frac{1}{2}\sin\left(\left(a+b\right)x\right)$

\cos (a x) \cos (b x) = \frac{1}{2} \cos ((a - b) x) + \frac{1}{2} \cos ((a + b) x)

$\cos\left(ax\right)\cos\left(bx\right)=\frac{1}{2}\cos\left(\left(a-b\right)x\right)+\frac{1}{2}\cos\left(\left(a+b\right)x\right)$

These formulas may be derived from the sum-of-angle formulas for sine and cosine.

Example: Evaluating $\displaystyle\int \sin\left(ax\right)\cos\left(bx\right)dx$

Evaluate ${\displaystyle\int}\sin\left(5x\right)\cos\left(3x\right)dx$ .

Show Solution

Apply the identity $\sin\left(5x\right)\cos\left(3x\right)=\frac{1}{2}\sin\left(2x\right)+\frac{1}{2}\sin\left(8x\right)$ . Thus,

\begin{array}{cc} \int \sin (5 x) \cos (3 x) d x & = \int \frac{1}{2} \sin (2 x) + \frac{1}{2} \sin (8 x) d x \\ = - \frac{1}{4} \cos (2 x) - \frac{1}{16} \cos (8 x) + C . \end{array}

$\begin{array}{cc}{\displaystyle\int}\sin\left(5x\right)\cos\left(3x\right)dx\hfill & ={\displaystyle\int}\frac{1}{2}\sin\left(2x\right)+\frac{1}{2}\sin\left(8x\right)dx\hfill \\ \hfill & =-\frac{1}{4}\cos\left(2x\right)-\frac{1}{16}\cos\left(8x\right)+C.\hfill \end{array}$

try it

Evaluate ${\displaystyle\int}\cos\left(6x\right)\cos\left(5x\right)dx$ .

Hint

Substitute $\cos\left(6x\right)\cos\left(5x\right)=\frac{1}{2}\cos{x}+\frac{1}{2}\cos\left(11x\right)$ .

Show Solution

$\frac{1}{2}\sin{x}+\frac{1}{22}\sin\left(11x\right)+C$

Module 3: Techniques of Integration